Problem: The 2050 volunteers are organizing the "Run! Chase the Rising Sun" activity. Starting on Apr 25 at 7:30 am, runners will complete the 6km trail around the Yunqi town.

There are n+1 checkpoints on the trail. They are numbered by 0, 1, ..., n. A runner must start at checkpoint 0 and finish at checkpoint n. No checkpoint is skippable — he must run from checkpoint 0 to checkpoint 1, then from checkpoint 1 to checkpoint 2 and so on. Look at the picture in notes section for clarification.

Between any two adjacent checkpoints, there are m different paths to choose. For any 1≤ i≤ n, to run from checkpoint i-1 to checkpoint i, a runner can choose exactly one from the m possible paths. The length of the j-th path between checkpoint i-1 and i is b_{i,j} for any 1≤ j≤ m and 1≤ i≤ n.

To test the trail, we have m runners. Each runner must run from the checkpoint 0 to the checkpoint n once, visiting all the checkpoints. Every path between every pair of adjacent checkpoints needs to be ran by exactly one runner. If a runner chooses the path of length l_i between checkpoint i-1 and i (1≤ i≤ n), his tiredness is $$$min_{i=1}^n l_i,$$$ i. e. the minimum length of the paths he takes.

Please arrange the paths of the m runners to minimize the sum of tiredness of them.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10 000). Description of the test cases follows.

The first line of each test case contains two integers n and m (1 ≤ n,m ≤ 100).

The i-th of the next n lines contains m integers b_{i,1}, b_{i,2}, ..., b_{i,m} (1 ≤ b_{i,j} ≤ 10^9).

It is guaranteed that the sum of n⋅ m over all test cases does not exceed 10^4.

Output

For each test case, output n lines. The j-th number in the i-th line should contain the length of the path that runner j chooses to run from checkpoint i-1 to checkpoint i. There should be exactly m integers in the i-th line and these integers should form a permuatation of b_{i, 1}, ..., b_{i, m} for all 1≤ i≤ n.

If there are multiple answers, print any.

Example

Input


2
2 3
2 3 4
1 3 5
3 2
2 3
4 1
3 5


Output


2 3 4
5 3 1
2 3
4 1
3 5

Note

In the first case, the sum of tiredness is min(2,5) + min(3,3) + min(4,1) = 6.

<image>

In the second case, the sum of tiredness is min(2,4,3) + min(3,1,5) = 3.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
 
public class AisejkbHi {
	static class FastReader{ 
        BufferedReader br; 
        StringTokenizer st; 
        public FastReader(){br = new BufferedReader(new InputStreamReader(System.in));}
        String next(){while (st == null || !st.hasMoreElements()){try{st = new StringTokenizer(br.readLine());} 
        catch (IOException  e){e.printStackTrace();}}return st.nextToken();}
        int nextInt(){ return Integer.parseInt(next());}long nextLong(){return Long.parseLong(next());}double nextDouble(){return Double.parseDouble(next());}
        String nextLine(){String str = ""; try{str = br.readLine(); } catch (IOException e) {e.printStackTrace();} return str; } 
    }
	static FastReader sc = new FastReader();
	static int mod = (int)(1e9+7);
	static int n;
	static String dp[][][];
	public static void main (String[] args) {
		PrintWriter out = new PrintWriter(System.out);
		int t = 1;
    	t = sc.nextInt();
    	z : while(t-->0) {
    		int n = sc.nextInt();
    		int m = sc.nextInt();
    		int b[][] = new int[n][m];
    		for(int i=0;i<n;i++) {
    			for(int j=0;j<m;j++) {
    				b[i][j] = sc.nextInt();
    			}
    		}
    		for(int i=0;i<n;i++) {
    			Arrays.sort(b[i]);
    		}
    		
    		int answer[][] = new int[n][m];
    		answer = b;
    		boolean visited[][] = new boolean[n][m];
    		int cnt = 0;
    		while(cnt<m) {
    			int min = Integer.MAX_VALUE;
    			int x = -1, y = -1;
    			for(int i=0;i<n;i++) {
    				for(int j=0;j<m;j++) {
    					if(answer[i][j]<min && !visited[i][j]) {
    						min = answer[i][j];
    						x = i; y = j;
    					}
    				}
    			}
    			visited[x][cnt] = true;
    			int temp = answer[x][cnt];
    			answer[x][cnt] = answer[x][y];
    			answer[x][y] = temp;
    			cnt++;
    		}
    		for(int i=0;i<n;i++) {
				for(int j=0;j<m;j++) {
					System.out.print(answer[i][j]+" ");
				}
				System.out.println();
			}
    	}
		out.close();
	}

	static int accumulate(int arr[], int first,int last)
	{
		int init = 0;
		for (int i = first; i< last; i++) {
			init = init + arr[i];
		}
		return init;
	}
	
	static Boolean isPossible(int arr[], int n)
	{
		if (n % 2 != 0)
		return false;
		int sum = accumulate(arr, 0, n);
		if (sum % 2 != 0)
		return false;
		
		int half[] = new int[n/2];
		
		return combinationUtil(arr, half, 0, n - 1,0, n, sum);
	}
	
	static Boolean combinationUtil(int arr[], int half[], 
	          int start, int end,
	          int index, int n,
	          int sum)
	{
		if (index == n / 2) {
		int curr_sum = accumulate(half, 0 , n/2);
		return (curr_sum + curr_sum == sum);
		}
		for (int i = start; i <= end && end - i + 1 >= 
		         n/2 - index; i++) {
		half[index] = arr[i];
		if (combinationUtil(arr, half, i + 1, end, 
		        index + 1, n, sum))
		return true;
		}
		
		return false;
	}
	private static int find(int n) {
		int sum = 0;
		while(n>0) {
			n/=2;
			sum++;
		}
		return sum;
	}
	static int ceil(int a,int b) {
		return a/b + (a%b==0?0:1);
	}
	static boolean prime[] = new boolean[2000009];
	static int fac[] = new int[2000009];
	
	static void sieve() {
		prime[0] = true;
		prime[1] = true;
		int max = 1000000;
		for(int i=2;i*i<=max;i++) {
			if(!prime[i]) {
				for(int j=i*i;j<=max;j+=i) {
					prime[j] = true;
					fac[j] = i;
				}
			}
		}
	}
	static int gcd(int a,int b) { if(b==0) return a; return gcd(b,a%b); }
}
 
class pair{
	int a;
	long b;
	pair(int a,long b){
		this.a = a;
		this.b = b;
	}
}